# Remove low orders from Series [closed]

It is easy to truncate Series upto some order, say $n$. My question is how do I remove low orders? Let us say my series is a power series in $x$. I want to remove the terms with negative powers because they diverge at $x = 0$. I can simply write

s1-s2, where

s1=Normal[Series[blah, {x, 0, n}]

s2=Normal[Series[blah, {x, 0, -1}]

but Mathematica does not understand to cancel the removed terms because they are complicated. The solution would be to use Collect[s1-s2, x, Simplify], but this is horribly slow as I increase $n$ above even 2. I suppose I could simply delete the terms by hand, but the outputs are very messy, and there must exist a proper way to do this.

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• It's going to be hard to guess without an example that reproduces the problem. Well, obviously, given 4 answers already, it's not really hard to guess, but it's hard to test whether one's guess is at all helpful. Another guess: Don't Normal the series and try s1 - Normal@s2. That might force it to collect like terms but only simplify the low-order ones. – Michael E2 Jul 16 '15 at 6:00

Why not to subtract two expansions like in

t1 = Series[1/Sin[x], {x, 0, 10}]
t2 = Series[1/Sin[x], {x, 0, 0}]

Then

Normal[t1] - Normal[t2]

Out[3]:= x/6 + (7 x^3)/360 + (31 x^5)/15120 + (127 x^7)/604800 + ( 73 x^9)/3421440

• This one is my favorite. – J. M.'s torpor Jul 16 '15 at 6:02
• @Guesswhoitis. This is the one the OP says does not work. – Michael E2 Jul 16 '15 at 6:08
• @MichaelE2 That's a good reason for choosing a favorite :) – Dr. belisarius Jul 16 '15 at 6:12
• @Michael, well damn, I would have gone without the Collect[] myself. I really hope OP would actually tell us the actual series he has… – J. M.'s torpor Jul 16 '15 at 6:22

I'm not sure this approach is applicable to all series, but from a quick test it seems to work for rational exponents:

Looking at the FullForm of

ser = Series[Exp[x]/x^(2/3), {x, 0, 5}]
(* x^(-2/3) + x^(1/3) + x^(4/3)/2 + x^(7/3)/6 + x^(10/3)/24 + x^(13/3)/120 + O[x]^(16/3) *)

gives

FullForm[ser]
(* SeriesData[x,0,List[1,0,0,1,0,0,Rational[1,2],0,0,Rational[1,6],0,0,Rational[1,24],0,0,Rational[1,120]],-2,16,3]

As we see, [[3]] contains a list of coefficients, while the lowest and highest powers are given by [[4]]/[[6]] and [[5]]/[[6]] respectively. If we want to eliminate all negative powers we may simply remove the -[[4]] first coefficients from the coefficient list, and set [[4]] to 0 afterwards. That is:

ser2 = ReplacePart[ser, {3 -> Drop[ser[[3]], -ser[[4]]], 4 -> 0}]
(* x^(1/3) + x^(4/3)/2 + x^(7/3)/6 + x^(10/3)/24 + x^(13/3)/120 + O[x]^(16/3) *)

If everything else fails, you can always do

Total[SeriesCoefficient[f@x, {x, 0, #}] x^# & /@ Range[0, 10]]

Is this as simple as

DeleteCases[s1, _*x^c_ /; c<0]

That is going to find all the terms in your series with negative exponents and simply delete them.